In sci.physics.relativity Tom S. wrote:

[...]
In the *exterior* region, the horizon seems to be located at a certain
*place* (namely r = 2m, where r is a space-like variable) and it sits at
that place for all eternity (all t).

Not quite. r is a spatial coordinate when r2m, but not at the horizon
itself.

In the *interior* region, the horizon seems to be located at a certain
instant of *time* (namely r = 2m, where r is now time-like) and it
exists at that instant 'everywhere' in the interior (i.e., for all t,
where t is spacelike).

Again, r is a "time" when r2m, but not at the horizon.

So what's left? The answer is that r=2m is a lightlike surface. The
horizon is moving outward at the speed of light -- but in a spacetime
that is so curved that the area of the horizon remains constant.

[...]
Thus, I am confused about whether or not an infalling observer would say
that she is *spatially* near the horizon for r just an itty bit smaller
than 2m. Does it make any sense to say that she is *spatially* closer
to the horizon when she arrives at r = 1.999m as compared to when she is
at r = 1.998m? Or is it not valid to ask about the spatial nearness of
events that are inside the horizon?

Spatial closeness is a bit tricky. To define it, you need a "time slice"
-- that is, you need to say at what time you are measuring the spatial
distance. Outside the horizon, there's a natural (though not unique)
choice: you choose as your time coordinate the "time" with respect to
which the black hole is not changing. Inside, no such time coordinate
exists.

Anyhow, though: inside the horizon, if you're using Schwarzschild
coordinates, r is a time, and t is a spatial coordinate. So the
answer to your question depends on where your observer is, that is,
on her value of t. (It's not *that* different from the outside. If
I ask, "Is the observer near the horizon at t=0?", you're answer is
presumably, "It depends on where she is.")

Steve Carlip

wrote:
LEJ Brouwer wrote:

JanPB wrote:
I think the manifold can be glued smoothly (one can write down a smooth
atlas on the quotient manifold) - it's the metric that has a "crease"
at the glued horizon. I posted an outline of an argument few minutes
ago. The problem is essentially with the Kruskal-Szekeres function
r(T,X) (the one given by the usual implicit equation) - it has a
nonzero slope at the horizon which would "crease" the metric there when
glued.

I agree with your analysis, and am no longer happy with the
implications of my original proposal. There was another possibility
which I had been trying rather to avoid (with hindsight maybe I
shouldn't have), in that the reflection of the infalling particle could
occur not at the event horizon, but rather at the central singularity -
this would also give rise to the same predictions, and I guess will not
upset the cart as the standard Schwarzschild interior solution remains
intact.

Has the possibility that a particle hitting the singularity is
reflected off of it both in space and time (i.e. into the white hole
interior solution) been suggested before that you know of? Are there
any good books or (esp. survey) articles discussing what might happen
(classically) at the Schwarzschild singularity to an infalling
particle?

I'm not sure that it makes sense to talk about the behavior of a classical
particle at the singularity, where tidal forces are infinite.

I think it does make good sense to talk about it,
as it seems to be an example of the clash between
classical and wave mechanical thinking.
From the standpoint of classical thinking we have
an infinite energy density at the location of the point
particle, and that renders an infinite curvature.
However, if we move 1 Planck length from that point,
the energy density via T_uv =0 when the volume used
to calculate the energy density is an infinitesmal dV.
From the standpoint of wave mechanics, there exists
a definite probability that some of that particle exists
at the 1 Planck = and as a result, the energy density
at one Planck is NOT zero and the energy density at
the particle point is NOT infinite.

That works well to combine how G_uv=T_uv may be
considered a probability of a particle within dV, IOW's
an energy density, where probability of a particle and
energy density are the same thing.
If so, wave mechanics is a GR solution, compatible
and quite straight forward.
Regards
Ken S. Tucker



















But what
you are talking about now might be related to a quotient of the Kruskal
extension described by Gibbons in "The Elliptic Interpretation Of Black
Holes And Quantum Mechanics," Nucl. Phys. B271 (1986) 497. (See also
Chamblin and Gibbons, gr-qc/9607079.) In this geometry, the two exterior
regions are identified, as are the two interior regions. You lose time
orientability, though -- there's no global definition of a past and future
direction.

There's another "single exterior" identification one can make as well,
the RP^3 geon. The best introduction I know of off hand is Louko,
gr-qc/9906031.

Steve Carlip

|
| Yes, of course.
| Some people even claim acceleration isn't real.
|
| Androcles
|
| Who are those people? HH

Pay me US$1000 and I'll tell you, ****head.
Androcles

I don't pay, because certainly you cannot find any adult person who does't
know what acceleration means. But there are a lot of people who don't
understand relative acceleration, including ....

Henry Haapalainen

"Henry Haapalainen" wrote in message
.. .
| |
| | Yes, of course.
| | Some people even claim acceleration isn't real.
| |
| | Androcles
| |
| | Who are those people? HH
|
| Pay me US$1000 and I'll tell you, ****head.
| Androcles
|
| I don't pay, because certainly you cannot find any adult person who does't
| know what acceleration means. But there are a lot of people who don't
| understand relative acceleration, including ....
|
| Henry Haapalainen
|
You don't pay because you never intended too, your reputation is ****.

wrote:
LEJ Brouwer wrote:

Has the possibility that a particle hitting the singularity is
reflected off of it both in space and time (i.e. into the white hole
interior solution) been suggested before that you know of? Are there
any good books or (esp. survey) articles discussing what might happen
(classically) at the Schwarzschild singularity to an infalling
particle?

I'm not sure that it makes sense to talk about the behavior of a classical
particle at the singularity, where tidal forces are infinite. But what
you are talking about now might be related to a quotient of the Kruskal
extension described by Gibbons in "The Elliptic Interpretation Of Black
Holes And Quantum Mechanics," Nucl. Phys. B271 (1986) 497. (See also
Chamblin and Gibbons, gr-qc/9607079.) In this geometry, the two exterior
regions are identified, as are the two interior regions. You lose time
orientability, though -- there's no global definition of a past and future
direction.

There's another "single exterior" identification one can make as well,
the RP^3 geon. The best introduction I know of off hand is Louko,
gr-qc/9906031.

Steve Carlip

Yes, these look very close to what I have in mind - except that I would
prefer not to have to actually _identify_ the two regions (what would
be the physical motivation for such an identification?), but rather to
'sew together' the regions at the singularity (which presumably must be
blown up a little to resolve it) and then superimpose, so rather than
the regions becoming two 'sides' of a single nonorientable manifold,
they correspond to two superimposed 'sheets' which happen to be
indistinguishable, so that the (hopefully) the property of
non-orientability, which for my purposes is a positive thing, remains.
The resulting manifold would then be eight-dimensional, and not four. I
don't particularly care whether the resulting manifold can support
spinors, as the nonorientability of the resulting will, according to
Hadley, endow the black hole itself with spinorial properties.

- Sabbir.

Tom S. wrote:

I don't think I've managed to ask my question very well. This is
undoubtedly due to my confusion. I am trying to wrap my feeble mind
around the notion of whether or not it makes any sense to ask, ''Where
in *space* is the horizon from the point of view of the interior Schw.
coordinates.''

In the *exterior* region, the horizon seems to be located at a certain
*place* (namely r = 2m, where r is a space-like variable) and it sits at
that place for all eternity (all t).

In the *interior* region, the horizon seems to be located at a certain
instant of *time* (namely r = 2m, where r is now time-like) and it
exists at that instant 'everywhere' in the interior (i.e., for all t,
where t is spacelike). [This is where my thinking is probably most off
base and where I need the most help.]

I think Schwarzschild's coordinates are just not really reasonable,
just like the Mercator map is not reasonable for the North Pole. They
come to us thanks to a set of certain assumptions designed to make
their derivation easier but these assumptions turn out quite
unreasonable from the physical point of view. Remember those pictures
of the black hole collapse they usually have in textbooks? These
pictures use the much more reasonable Eddington-Finkelstein chart which
keeps the r coordinate unchanged but adjusts the time in a way that
makes r direction spacelike throughout and intuitively sensible. The
light cones do not suddenly flip at the horiozon but gradually tip
towards the singularity (at the horizon they become tangent to it). So
one can think about r=1.9m or r=2.1m, etc.

It's somewhat interesting to see the spaces of Eddington-Finkelstein
constant time on the Kruskal-Szekeres diagram - I plotted a few of
those he
http://www.mastersofcinema.org/jan/efks.gif (m=1.0)
http://www.mastersofcinema.org/jan/efks1.gif (same thing with slice
t=0 highlighted)
http://www.mastersofcinema.org/jan/efks2.gif (m=3.0)

--
Jan Bielawski

wrote in message
...

In sci.physics.relativity Tom S. wrote:
[...]

r is a spatial coordinate when r2m, but not at the horizon
itself. [...] r is a "time" when r2m, but not at the horizon.
So what's left? The answer is that r=2m is a lightlike surface. The
horizon is moving outward at the speed of light -- but in a spacetime
that is so curved that the area of the horizon remains constant.


Moving outward at the speed of light relative to any local inertial frame
that happens to be 'momentarily falling through' the horizon, I reckon. OK,
that helps.

[...]
Does it make any sense to say that [an infalling observer] is
*spatially* closer to the horizon when she arrives at
r = 1.999m as compared to when she is at r = 1.998m?

Spatial closeness is a bit tricky. To define it, you need a "time slice"
-- that is, you need to say at what time you are measuring the spatial
distance. [...] Inside, no such time coordinate exists.


I'm getting the feeling that my question is rather meaningless.

.inside the horizon, if you're using Schwarzschild
coordinates, r is a time, and t is a spatial coordinate. So the
answer to your question depends on where your observer is, that is,
on her value of t.

If the answer depends on t, then what specific values of t in the interior
region would correspond to events 'near' the horizon? t going to infinity?
[Sorry to keep pushing for a specific answer to this (bad?) question.]
Somewhat related to this is the question:

Suppose the infalling observer begins her journey in the exterior region 'at
rest' at r = 3m, say, and t = 0. Does this determine a unique value for t in
the interior region when the observer arrives at r = 1.9m, say? If not,
then I don't see how a value of t for an event in the interior has much
relevance to deciding if the event is 'spatially near' the horizon.

Anyway, I think you've helped me see that 'spatial nearness to the horizon'
in the interior is probably not meaningful. Thanks.

Tom S.

Tom S. wrote:
wrote in message
...
The
horizon is moving outward at the speed of light -- but in a spacetime
that is so curved that the area of the horizon remains constant.

Moving outward at the speed of light relative to any local inertial
frame that happens to be 'momentarily falling through' the horizon, I
reckon.

Yes.

So, for instance, if a spaceship is repeatedly sending light pulses from
nose to tail, once its nose enters the horizon pulses from nose to tail
do not cross the horizon, because the tail will cross the horizon before
the pulse can reach it.

If the answer depends on t, then what specific values of t in the
interior region would correspond to events 'near' the horizon?

Any value of t will do. Just as on the surface of the earth, any spatial
position is "near" 1 second ago. Remember in the interior region, r is
timelike (-d/dr is future pointing for an infalling path), so the
horizon r=2M is "a short time ago" rather than some distance away.

You are probably used to thinking "I am moving at 10 m/sec relative to
the earth, so 1 second ago I was 10 meters back there". That works on
earth, but not near the horizon of a black hole -- specifically you had
the earth's surface for reference, but at the horizon it is not possible
for there to be any sort of spatial reference (any timelike object there
must be infalling).

t going
to infinity?

For an infalling timelike _geodesic_, as the path approaches the
horizon, the exterior t goes to infinity. Immediately after the horizon
the internal t decreases rapidly from infinity as the geodesic falls
further into the interior.

Suppose the infalling observer begins her journey in the exterior region
'at rest' at r = 3m, say, and t = 0. Does this determine a unique value
for t in the interior region when the observer arrives at r = 1.9m,
say?

No.

If not, then I don't see how a value of t for an event in the
interior has much relevance to deciding if the event is 'spatially near'
the horizon.

It doesn't. r is the relevant coordinate for being "near" the horizon,
but in the interior it is timelike and the horizon is "some time ago"
rather than in any particular direction in space. So, too, the
singularity at r=0 is "some time in the future" and not in any
particular direction, either.

Tom Roberts

"JanPB" wrote in message
ups.com...

Tom S. wrote:
In the *exterior* region, the horizon seems to be located at a certain
*place* (namely r = 2m, where r is a space-like variable) and it sits at
that place for all eternity (all t).

In the *interior* region, the horizon seems to be located at a certain
instant of *time* (namely r = 2m, where r is now time-like) and it
exists at that instant 'everywhere' in the interior (i.e., for all t,
where t is spacelike). [This is where my thinking is probably most off
base and where I need the most help.]

I think Schwarzschild's coordinates are just not really reasonable,
just like the Mercator map is not reasonable for the North Pole. They
come to us thanks to a set of certain assumptions designed to make
their derivation easier but these assumptions turn out quite
unreasonable from the physical point of view. Remember those pictures
of the black hole collapse they usually have in textbooks? These
pictures use the much more reasonable Eddington-Finkelstein chart which
keeps the r coordinate unchanged but adjusts the time in a way that
makes r direction spacelike throughout and intuitively sensible.

Thanks, Jan. That's something that I hadn't realized (but should have, of
course). Without changing the r-labels of events you can make the
r-direction spacelike in both the interior and exterior regions by just
replacing the Schw-t-labels of the events with the EF-time-labels. A
'moment of Zen' for me.

The
light cones do not suddenly flip at the horizon but gradually tip
towards the singularity (at the horizon they become tangent to it). So
one can think about r=1.9m or r=2.1m, etc.

Yes, good.

It's somewhat interesting to see the spaces of Eddington-Finkelstein
constant time on the Kruskal-Szekeres diagram - I plotted a few of
those he
http://www.mastersofcinema.org/jan/efks.gif (m=1.0)
http://www.mastersofcinema.org/jan/efks1.gif (same thing with slice
t=0 highlighted)
http://www.mastersofcinema.org/jan/efks2.gif (m=3.0)

Thanks for taking the time to do this.
Tom S.